Math 180, Exam 1, Study Guide
Problem 1 Solution
1. What is the slope of the linear function f (x) whose graph goes through the points (1, 2)
and (4, 4)? What is the value of f (7)?
Solution: The slope of the line that goes through the points (1, 2) and (4, 4) is:
m=
y2 − y1
4−2
2
=
=
x2 − x1
4−1
3
Using the slope above and the point (1, 2), we write an equation for the line as follows:
y − y1 = m(x − x1 )
2
y − 2 = (x − 1)
3
2
2
y−2= x−
3
3
4
2
y = x+
3
3
4
2
Therefore, the linear function is f (x) = x + . The value of f (7) is:
3
3
2
4
f (7) = (7) + = 6
3
3
1
Math 180, Exam 1, Study Guide
Problem 2 Solution
2. Find lim
x→2
x2 − 3x + 2
.
x2 − x − 2
Solution: When substituting x = 2 into the function f (x) =
x2 − 3x + 2
we find that
x2 − x − 2
x2 − 3x + 2
22 − 3(2) + 2
0
=
=
2
2
x −x−2
2 −2−2
0
which is indeterminate. We can resolve the indeterminacy by factoring.
lim
x→2
x2 − 3x + 2
(x − 1)(x − 2)
= lim
2
x→2 (x + 1)(x − 2)
x −x−2
x−1
= lim
x→2 x + 1
2−1
=
2+1
=
1
1
3
Math 180, Exam 1, Study Guide
Problem 3 Solution
3. The graph of function f (x) is below. At what point or points is f (x) not continuous?
3
2
1
0
0
1
2
3
4
5
6
7
Solution: The function is not continuous at x = 4 because the one-sided limits
lim f (x) = 2
lim f (x) = 1,
x→4+
x→4−
are not equal to each other.
1
Math 180, Exam 1, Study Guide
Problem 4 Solution
4. Find the derivatives of the following functions using the basic rules. Do not simplify your
answer.
(a) x3 + x1/3 ,
(b) x2 ex ,
(c)
2+x
3 + x2
Solution:
(a) Use the Power Rule.
1
(x3 + x1/3 )′ = 3x2 + x−2/3
3
(b) Use the Product Rule.
(x2 ex )′ = x2 (ex )′ + (x2 )′ ex
= x2 ex + 2xex
(c) Use the Quotient Rule.
′
2+x
(3 + x2 )(2 + x)′ − (2 + x)(3 + x2 )′
=
3 + x2
(3 + x2 )2
=
(3 + x2 ) − (2 + x)(2x)
(3 + x2 )2
1
Math 180, Exam 1, Study Guide
Problem 5 Solution
5. Find the derivatives of the following functions using the basic rules. Do not simplify your
answer.
1
1
(a) 1 + x + x2 + x3 ,
2
6
(b) (x − 1)ex ,
(c) √
1
x−1
Solution:
(a) Use the Power Rule.
′
1 2 1 3
1
1+x+ x + x
= 1 + x + x2
2
6
2
(b) Use the Product Rule.
[(x − 1)ex ]′ = (x − 1)(ex )′ + (x − 1)′ ex
= (x − 1)ex + ex
(c) Use the Quotient Rule.
1
√
x−1
′
√
√
( x − 1)(1)′ − (1)( x − 1)′
√
=
( x − 1)2
0 − 2√1 x
√
=
( x − 1)2
1
Math 180, Exam 1, Study Guide
Problem 6 Solution
6. Find the equation of the tangent line to y = f (x) at x = 2 for the function f (x) = −x2 +7x.
Do not simplify your answer.
Solution: The derivative f ′ (x) is found using the Power Rule.
f ′ (x) = (−x2 + 7x)′ = −2x + 7
At x = 2 the values of f and f ′ are:
f (2) = −22 + 7(2) = 10
f ′ (2) = −2(2) + 7 = 3
We now know that the point (2, 10) is on the tangent line and that the slope of the tangent
line is 3. Therefore, an equation for the tangent line in point-slope form is:
y − 10 = 3(x − 2)
1
Math 180, Exam 1, Study Guide
Problem 7 Solution
7. Let f (x) = x2 + 3.
(a) Find the average rate of change of f (x) over the interval 1 ≤ x ≤ 3.
(b) Find the instantaneous rate of change of f (x) at x = 2.
Solution:
(a) The average rate of change formula is:
average ROC =
f (b) − f (a)
b−a
Using f (x) = x2 + 3, b = 3, and a = 1 we have:
average ROC =
(32 + 3) − (12 + 3)
= 4
3−1
(b) The instantaneous rate of change at x = 2 is f ′ (2). The derivative f ′ (x) is:
f ′ (x) = 2x
At x = 2 we have:
instantaneous ROC = f ′ (2) = 2(2) = 4
1
Math 180, Exam 1, Study Guide
Problem 8 Solution
8. Let f (x) = x2 . Use the definition of the derivative as the limit of the difference quotient
to find f ′ (3).
Solution: There are two possible difference quotients we can use to evaluate f ′ (3). One is:
f ′ (3) = lim
h→0
f (h + 3) − f (3)
(h + 3)2 − 32
= lim
.
h→0
h
h
The other is:
f ′ (3) = lim
x→3
x2 − 32
f (x) − f (3)
= lim
x→3 x − 3
x−3
Evaluating the first limit above we have:
(h + 3)2 − 32
h→0
h
h2 + 6h + 9 − 9
= lim
h→0
h
2
h + 6h
= lim
h→0
h
= lim (h + 6)
f ′ (3) = lim
h→0
=0+6
= 6
Evaluating the second limit we have:
x2 − 32
f (3) = lim
x→3 x − 3
x2 − 9
= lim
x→3 x − 3
(x + 3)(x − 3)
= lim
x→3
x−3
= lim (x + 3)
′
x→3
= 3+3
= 6
1
Math 180, Exam 1, Study Guide
Problem 9 Solution
9. Use the table below, which shows values of f (x) for x near 2,
x
f (x)
1.8 1.9 2.0 2.1 2.2
2.24 2.27 2.30 2.33 2.37
to find the slope of a secant line that is an estimate for f ′ (2).
Solution: An approximate value for f ′ (2) is
f ′ (2) ≈
2.33 − 2.30
f (2.1) − f (2.0)
=
= 0.3
2.1 − 2.0
0.1
1
Math 180, Exam 1, Study Guide
Problem 10 Solution
√
10. Find lim
x→1
√
3x + 1 − 2x + 2
.
x−1
√
Solution: When substituting x = 1 into the function f (x) =
that
√
3x + 1 − 2x + 1
, we find
x−1
p
p
√
3(1) + 1 − 2(1) + 2
3x + 1 − 2x + 2
0
=
=
x−1
1−1
0
which is indeterminate. We can resolve the indeterminacy by multiplying f (x) by the “conjugate” of the numerator divided by itself.
√
√
√
√
√
√
3x + 1 − 2x + 2
3x + 1 − 2x + 1
3x + 1 + 2x + 2
√
lim
= lim
·√
x→1
x→1
x−1
x−1
3x + 1 + 2x + 2
(3x + 1) − (2x + 2)
√
√
= lim
x→1 (x − 1)( 3x + 1 +
2x + 2)
(x − 1)
√
√
= lim
x→1 (x − 1)( 3x + 1 +
2x + 2)
1
√
= lim √
x→1
3x + 1 + 2x + 2
1
p
=p
3(1) + 1 + 2(2) + 2
√
=
1
4
1
√
.
3x + 1 + 2x + 2
This is possible because the function is continuous at x = 1. In fact, the function is contin1
uous at all x ≥ − .
3
We evaluated the limit above by substituting x = 1 into the function √
1